How do you find the derivative of #y=tan(3x)# ?

Question

Aurora Alvarado · Accepted Answer

Since there is a function inside a function (a "composite" function), you could use the chain rule to accomplish this. It is as follows: The derivative of a composite function F(x) is as follows: #F'(x)=f'(g(x)) (g'(x))# Alternatively put: = the derivative of the inner function multiplied by the derivative of the outer function when the inner function is left unaltered. So let's look at your question.
#y=tan (3x)# The outer function is tan and the inner function is #3x#, since #3x# is "inside" the tan. Think of it as #tan(u)# where #u=3x#, so the #3x# is composed in the tan. Deriving, we get: The derivative of the outer function (leaving the inside function alone):
#d/dx tan(3x)=sec^2(3x)#
The derivative of the inner function:
#d/dx 3x=3#
Combining, we get:
#d/dx y=y'= 3sec^2(3x)#

Elijah Abram · Answer

undefined To find the derivative of $y = \tan(3x)$, you'll use the chain rule, which is a fundamental technique in calculus for differentiating composite functions. The chain rule states that if you have a composite function $f(g(x))$, then its derivative can be found by multiplying the derivative of the outer function by the derivative of the inner function. The derivative of $\tan(x)$ is $\sec^2(x)$, and the derivative of $3x$ with respect to $x$ is 3.

So, applying the chain rule to $y = \tan(3x)$, you get:

$$ \frac{dy}{dx} = \sec^2(3x) \times \frac{d}{dx}(3x) $$

$$ \frac{dy}{dx} = 3\sec^2(3x) $$

Therefore, the derivative of $y = \tan(3x)$ with respect to $x$ is $3\sec^2(3x)$.